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show that cos(3theta) = cos^3(theta)-3 cos(theta)sin^2(theta) ... use this to find an equation in rectangular coordinates for the curve r= 4 cos (3theta)...... ANY HELP ?
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It's a bit of arithemetic, but you'll need the formulas for sin(a + b) and cos (a + b). Not sure if you're allowed to use the double angle formulas, so we'll do it the long way First, the angle sum forumlas: \[sin(\alpha + \beta) = sin\alpha cos\beta + sin\beta cos\alpha\]\[cos(\alpha + \beta)=cos\alpha cos\beta - sin\alpha sin\beta\]
So, \[cos(3\theta)=cos(2\theta+\theta)=cos(2\theta)cos\theta-sin(2\theta)sin\theta\]\[=(cos^{2}\theta-sin^{2}\theta)cos\theta - (2sin\theta cos\theta)sin\theta\] From there, it's distribute and simplify...
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